An ester (A) (C_4H_8O_2) , on treatment with excess methyl magnesium chloride followed by acidification, gives an alcohol (B) as the sole organic product. Alcohol (B), on oxidation with NaOCI followed by acidification gives acetic acid. Deduce the structures of (A) and (B). show the reactions involved.

An ester (A) (C_4H_8O_2) , on treatment with excess methyl magnesium c

1.	An ester (A) (C_4H_8O_2) , on treatment with excess methyl magnesium chloride followed by acidification, gives an alcohol (B) as the sole organic product. Alcohol (B), on oxidation with NaOCI followed by acidification gives acetic acid. Deduce the structures of (A) and (B). show the reactions involved.
Answer» Solution :Let the ester be `R-overset(O)overset(\|\|)C-OR`', i.e., `C_4H_8O_2`. The reactions of an ester with methtyl magnesium chloride are as follows : `R-overset(O)overset(\|\|)underset((A))C-OR'overset(CH_3MgCI)toR-overset(OMgCI)overset(\|)underset(CH_3)underset(\|)C-OR'overset(H^+)to R'OH+R-overset(O)overset(\|\|)C-CH_3overset(CH_3MgCI)toR-overset(OMgCI)overset(\|)underset(CH_3)underset(\|)C-CH_3overset(H^+)toR-overset(OH)overset(\|)underset((B))underset(CH_3)underset(\|)C-CH_(3)overset([O])underset(NaOCI)toCH_3COOH` THUS, R, should be H, because only secondary alcohol can be oxidised by `NaOCI` and not tertiary. Hence, the ester is `H-overset(O)overset(\|)C-OC_3H_7(C_4H_8O_2)` and R'OH on oxidation GIVES `CH_3COOH`. Thus, R'OH must be secondary alcohol of three carbon ATOMS, `(CH_3)_2CH-OH`, i.e., R' is isoproyl `(CH_3)_2CH-`. Therefore, the ester (A) is `H-overset(O)overset(\|\|)C-O-overset(H)overset(\|)underset(CH_3)underset(\|)C-CH_3` (isopropyl formate) and alcohol (B) is a secondary alcohol , `CH_3-underset(OH)underset(\|)CH-CH_3` (isopropyl alcohol ).